As x -> infinity we have the same behavior for 1/x so the function again has a horizontal asymptote at y = 2 for x->infinity.

(Technically there are still two horizontal asymptotes, even though they are the same line.)

-Dan

Sep 28th 2006, 06:14 AM

topsquark

Quote:

Originally Posted by tmd

f(x) = { sin x/x, x cant be 0
k, x = 0
In order for f(x) to be continous at x=0, the value of k must be ?

Are you saying that for x not equal to zero f(x) = sin(x)/x and at x = 0 f(x) = k?

For f(x) to be continous we must have that k = lim(x->0) sin(x)/x. There are various ways to find this limit. Do you know the power series expansion of sin(x)? If so, this is probably the simplest way to get this limit.

For x close to 0
sin(x) = x - (1/3!)x^3 + (1/5!)x^5 - ...

So for x close to 0
sin(x)/x is approximately [x - (1/3!)x^3 + (1/5!)x^5 - ...]/x = 1 - (1/3!)x^2 + (1/5!)x^4 - ...
which is approximately 1. ( x^n << 1 for x close to 0 for all positive integers n)

No need to do that.
The limit of,
sin(x)/x
Is one of those students must know immediately.

Sep 28th 2006, 07:41 AM

ThePerfectHacker

Quote:

Originally Posted by tmd

1. (2 + e^1/x) / (2 - e^1/x)
What is the horizontal asymptotes?

Divide the numerator and demoninator by e^(1/x)
Thus,
(2e^(-1/x)+1)/(2e^(-1/x)-1)
When x---> +oo
The numerator is, 2(1)+1=3
The denominator is, 2(1)-1=1 not equal to zero.
Thus, by the quotient rule of limits this limit is, (3)/(1)=3
That is one horizontal asymptote.
---
Instead of dividing leave it the way it should be
(2 + e^(1/x)) / (2 - e^(1/x))
As x---> -oo
The numerator is 2+1=3
The denominator is 2-1=1 not equal to zero.
Thus, by the qtuotient rule of limits this limit is (3)/(1)=3
Also the same.